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We are given a set of natural numbers 2, 3, 4, ..., n. Consider all subsets, each of them consisting of the combinations $_{(n-1)}C_{2}$, $_{(n-1)}C_{3}$, $_{(n-1)}C_{4}$ etc. We take the products of the terms in each such subset and then its reciprocals. Find the sum of all these reciprocals.

I believe I must find a recursive formula but I have no idea how to proceed...

I tried for n=4 and n=5 and found 5/12 and 86/120 respectively but I don't know how to continue. For example, for n=4: Let's consider the set {2,3,4}. Then we have the subsets {2,3}, (2,4), (3,4), (2,3,4) and their respective elements' products are 6, 8, 12 and 24. Then we take $\frac{1}{6}, \frac{1}{8}, \frac{1}{12}, \frac{1}{24}$ and add them. The result is $\frac{10}{24}$.

Also noticed that for each such fraction, the product of the numerator and the denominator equals n!.

Tom Galle
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  • @MartinR In that one, 1 is also included. – Tom Galle Jul 16 '19 at 07:53
  • Apologies, I meant for n=4 and n=5. – Tom Galle Jul 16 '19 at 07:55
  • Let's consider the set {2,3,4}. Then we have the subsets {2,3}, (2,4), (3,4), (2,3,4) and their respective elements' products are 6, 8, 12 and 24. Then we take $\frac{1}{6}, \frac{1}{8}, \frac{1}{12}, \frac{1}{24}$ and add them. – Tom Galle Jul 16 '19 at 08:01
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    You should place your example in the question itself rather than in the comments, where it might be overlooked. – N. F. Taussig Jul 16 '19 at 08:16
  • Your question is not clear or contradicts itself. It says consider all subsets, and then goes one listing a number of binomial coefficients "$\binom{n-1}2,\binom{n-1}3,\binom{n-1}4$ etc.", from which notably $\binom{n-1}0$ and $\binom{n-1}1$ appear to be omitted (but it does not say so, and the "etc." might mean they are included, just not deemed sufficiently interesting to merit explicit mention). But subsets are not binomial coefficients, they are just counted by binomial coefficients, and the rest of the question suggests it is the subsets, not binomial coefficients, that are handled. – Marc van Leeuwen Jul 16 '19 at 09:46

1 Answers1

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Let $x_n$ denote your sum of the reciprocals of all products of subsets of $\{ 2, \ldots, n \}$ with at least 2 elements. Then $$ x_n + \left( \frac 12 + \ldots + \frac 1n \right) $$ is the sum of the reciprocals of all products of non-empty subsets of $\{ 2, \ldots, n \}$, and $$ \frac 11 + 2 x_n + 2 \left( \frac 12 + \ldots + \frac 1n \right) $$ is the sum of the reciprocals of all products of non-empty subsets of $\{ 1, \ldots, n \}$. According to Show that the sum of reciprocal products equals $n$ that sum is exactly $n$, therefore $$ x_n = \frac 12 (n-1) - \left( \frac 12 + \ldots + \frac 1n \right) = \frac 12 (n+1) - H_n $$ where $H_n$ is the $n^\text{th}$ harmonic number.

Martin R
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